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3 years ago

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A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th

at she traveled 0.1 mi while accelerating. Determine (a) the acceleration of the car, (b) the time required to reach 65 mi/h.

1 answer:

Helga [31]3 years ago

5 0

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

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The speed of a vehicle is reduced with a constant acceleration from 72km/h to 18

Radda [10]

Answer:

The correct answer will be "1477.84 N".

Explanation:

Given that,

Mass,

m = 1.6 mg

or,

= 1600 kg

Initial velocity,

u = 72 km/h

= $72\times \frac{5}{18} \ m/s$

= $20 \ m/s$

Final velocity,

v = 18 km/h

= $18\times \frac{5}{18}$

= $5 \ m/s$

Covered distance,

s = 250 m

By using the below relation, we get

⇒ $v^2=u^2+2as$

On putting the values, we get

⇒ $(5)^2=(20)^2+2\times a\times 250$

⇒ $a=-0.75 \ m/s^2$ (shows the deceleration)

Slope will be given as 1 in 25, then

⇒ $Sin \theta=\frac{1}{25}$

$\theta=2.3^{\circ}$

hence,

As we know,

⇒ $\Sigma F=ma$

or,

⇒ $Braking \ force+350-mgSin\theta=ma$

⇒ $Braking \ force=ma+mgSin\theta-350$

On substituting all the values, we get

⇒ $=1600(0.75+1600\times 9.81 Sin(2.3^{\circ})-350$

⇒ $=1477.84 \ N$

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3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

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